令f(x)＝lnx－(ex-2),则f‘(x)＝1/x－ex,令f‘(x)＝0,即1/x－ex＝0,解得x＝1/e；令f‘(x)＞0,即1/x－ex＞0,解得x∈(0,1/e)；令f‘(x)＜0,即1/x－ex＜0,解得x∈(1/e,+∞)；∴当x∈(0,1/e),f‘(x)＞0,f(x)单调递增；x...